Question

1. Father Mendel has found a new variant of peas to work with.  The true-breeding purple pea variety (pu) is mated to a true-breeding wrinkled (w) individual.  All of the F1 have yellow smooth peas

A) What phenotypes, and in what relative quantities, would Father Mendel expect to see among the F2 generation produced by crossing the F1 hybrids to individuals with purple wrinkled peas(i.e., this is a test cross, not a selfing of the F1)?  Draw the Punnett Square (or rectangle) illustrating why Mendel would expect to see these ratios of phenotypes among the offspring.

B. As a modern pea breeder, you know that the pu and w genes are linked, placed 12 cM apart on the same chromosome.  In this case, what phenotypes would you expect to see, and at what relative frequencies, among the offspring of the generation produced by crossing the F1 hybrids to individuals with purple wrinkled peas? Explain/diagram your reasoning.

Answer 1

Answer: Let the genotype of true breeding purple pea variety be pupu

And the genotype of true breeding wrinkled pea variety be ww

Purple pea variety is pure for smooth character hence the genotype of P1 will be pupu++ (+ indicates dominant variety)

And the genotype of wrinkled pea variety P2 which is pure for yellow color be ++ww

Hence the F1 offspring will have the genotype +pu+w and the phenotypes will be yellow and smooth.

The gamates produced by F1will be ++, +w, pu+, puw

These are test crossed with purple,wrinkled plants which will have the genotype pupuww which will produce only one type of gamate i.e puw

The cross can be shown as follows

gamates	++	+w	pu+	puw
puw	+pu+w Yellow,smooth	+puww Yellow,wrinkled	pupu+w Purple,smooth	pupuww Purple,wrinkled

Mendel would expect that in the F2 generation, all the offspring will be in equal proportion i.e the ratio would be 1:1:1:1

According to Mendel’s third law, the law of independent assortment in this dihybrid cross the characters of the color and texture are independent of each other and they will enter into gamates independent of each other giving all the possible combinations of the characters in equal proportion.

Answer b) With our current knowledge, if we know that the genes are on the same chromosome and they are just 12cM apart from each other, it can be said that the genes are linked and they will undergo recombination less frequently resulting more of parental combinations and less of recombinants.

If the distance between the genes is 12cM that means the recombination frequency is 12 %

The recombination frequency gives the number of recombinants, in this case referring to the above Punnett square the recombinants are

+puww (Yellow,wrinkled) and pupu+w (Purple,smooth)

The number of recombinants will be12 out of 100

Therefore the parental combinations +pu+w (Yellow,smooth) and pupuww (Purple,wrinkled) will be 88 out of 100

Hence the relative frequencies will be

44(yellow,smooth): 6(yellow, wrinkled):6(purple,smooth):44 (smooth,wrinkled)